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3 years ago

Match each simple machine with its description.

Physics

1 answer:

kipiarov [429]3 years ago

8 0

Answer:

inclined plane- 3

wheel and axel- 1

pulley- 5

lever- 4

wedge- 2

Explanation:

got it right on edge :)

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When the displacement in SHM is one-half the amplitude xm, what fraction of total energy is (a) kinetic energy and (b) potential

Svet_ta [14]

Answer:

Explanation:

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2 years ago

The slow, steady downhill flow of loose, weathered Earth materials is called A. flow. B. slide. C. creep. D. slump.

Art [367]

I beileve its b
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3 years ago

Read 2 more answers

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Debora [2.8K]

Answer:

1e , 2j , 3c , 4d , 5k , 6a , 7i , 8b , 9m , 10h ,11g , 12f , 13l .

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3 years ago

In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step

satela [25.4K]

Answer:

W = 16.5 Kj

P = 49.9 Watt

E = 16471

Explanation:

m = 73.5kg

t = 5min 30sec = (5×60) + 30 = 330sec

each step = 16.6cm = 0.166m

h = 135×0.166 = 22.41 m

g = 10 m/s²

(i) W = F × s = W × h = mgh

W = 73.5×10×22.41 = 16471.35

W = 16.5 Kj

(ii) Power = workdone/time

P = 16471.35/330

P = 49.9 Watt

(iii) The energy burnt in this process = 16471

4 0

3 years ago

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$

So, the energy lost by the meteorite corresponds to 0.110 megatons.

The energy of one atomic bomb explosion in Hiroshima is equal to

$E'=13 kt$ (13 kilotons)

which corresponds to

$E'=0.013 Mt$ (0.013 megatons)

Here the energy of the meteorite is equal to

$E=0.110 Mt$ (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

$\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46$

So, 8.46 bombs.

5 0

3 years ago